In solving certain types of equations, one may obtain besides the proper (right) roots also some strange roots which do not satisfy the original equation. Such a thing can happen especially when one has in some stage squared both sides of the treated equation; in this situation one must check all ``roots'' by substituting them to the original equation.

Example. $$x-\sqrt{x} = 12$$ $$x-12 = \sqrt{x}$$ $$(x-12)^2 = (\sqrt{x})^2$$ $$x^2-24x+144 = x$$ $$x^2-25x+144 = 0$$ $$x = \frac{25\pm\sqrt{25^2-4\cdot144}}{2} = \frac{25\pm7}{2}$$ $$x = 16 \quad \lor \quad x = 9$$ Substituting these values of $x$ into the left side of the original equation yields $$16-4 = 12, \quad 9-3 = 6.$$ Thus, only $x = 16$ , is valid, $x = 9$ , is a strange root. (How $x = 9$ , is related to the solved equation, is explained by that it may be written $(\sqrt{x})^2-\sqrt{x}-12 = 0$ from which one would obtain via the quadratic formula that $\sqrt{x} = \frac{1\pm7}{2}$ i.e. $\sqrt{x} = 4$ , or $\sqrt{x} = -3$ The latter corresponds the value $x = 9$ but it were relevant to the original equation only if we would allow negative values for square roots of positive numbers; the current practice excludes them.)

The general explanation of strange roots when squaring an equation is, that the two equations $$a = b,$$ $$a^2 = b^2$$ are not equivalent (but the equations $a = \pm b$ , and $a^2 = b^2$ , would be such ones).